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Let $X_1, X_2,...$ be i.i.d integrable random variables in $\mathbb{R}$ with $\mathbb{E}[X_i] =0$ and $\mathbb{P} (X_i >0) >0$. Let $x>0$, $S_0 = x$, and $S_n= x + \sum_{i=1}^{n} X_i $. For every $0<r< \infty$, we define \begin{equation} \eta = \inf \{ n \geq 0 : S_n \leq 0 \text{ or } S_n \geq r\}. \end{equation} It is not hard to see that $\mathbb{E} [ \eta ] < \infty$. However, the problem is to show that $X_{\eta}$ is integrable. (It is not trivial at all, as the process $\{X_n\}$ is NOT a martingale. Thus, Wald's Identity or Optional Stopping Theorem do not work.) Any ideas?

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  • $\begingroup$ The random variables are not necessarily square integrable, right? (Because in this case, the claim is rather obvious.) $\endgroup$ – saz Sep 5 '14 at 15:54
  • $\begingroup$ NO, unfortunately. $\endgroup$ – Richard Sep 5 '14 at 16:28
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If $x$ is not in $(0,r)$, then $\eta=0$ and $X_\eta$ is undefined, hence we assume that $x$ is in $(0,r)$.

Since one knows that $\eta$ is integrable, Wald's theorem ensures that $S_\eta$ is integrable (and provides its expectation, which we will not need). Furthermore, $S_{\eta-1}$ is in $(0,r)$ by the definition of $\eta$ and $|X_\eta|\leqslant|S_\eta|+|S_{\eta-1}|$ by the triangular inequality, hence $|X_\eta|\leqslant|S_\eta|+r$ almost surely, which proves that $X_\eta$ is integrable.

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First, $S_\eta$ is a martingale whose increments have finite conditional expectation and $\mathbb{E}[\eta] < \infty$, so the optional stopping theorem implies that $S_\eta$ is integrable. Second, notice that $\{S_\eta \leq 0\}=\{X_\eta \leq 0\}$. Together, these imply that $$-\infty < \mathbb{E}[S_\eta I(S_\eta \leq 0)]$$ $$\quad\quad\quad\quad\quad= \mathbb{E}[(S_{\eta-1} + X_\eta)I(X_\eta \leq 0)]$$ $$\quad\quad\quad\quad\leq \mathbb{E}(r+X_\eta)I(X_\eta \leq 0)]$$ $$\quad\quad\quad\quad\leq r + \mathbb{E}[X_\eta I(X_\eta \leq 0)]$$ where $I$ is the indicator function. This implies that $\mathbb{E}[X_\eta^-] <\infty$.

Try doing something similar starting with $\infty > \mathbb{E}[S_\eta I(S_\eta >0)]$ to conclude that $\mathbb{E}[X_\eta^+] < \infty$. Then, notice that we've failed to take into account the possibility that $\eta <2$ and make the necessary modifications.

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